Question

The light sensitive compound on most photographic films is silver bromide AgBr. A film is exposed when the light energy absorbed dissociates this molecule into its atoms. The energy of dissociation of AgBr is ${10}^{5}Jmo{l}^{-1}$. For a photon that is just able to dissociate a molecule of AgBr, the photon energy is

• A. 1.04 eV
• B. 2.08 eV
• C. 3.12 eV
• D. 4.16 eV

Answer 1

The correct option is A 1.04 eV

1 mole of AgBr contains $6.022\times {10}^{23}$ molecules and $1J$ of energy is equivalent to $6.24\times {10}^{18}eV$.
Thus, energy needed to dissociate 1 molecule of AgBr is given by:

$E=\frac{{10}^5\times 6.24\times {10}^{18}}{6.022\times {10}^{23}}=1.04eV$